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Use induction on the order of the finite p-group. The trivial group is nilpotent. For $G\ne 1$, the permitted center theorem gives $Z=Z(G)\ne 1$, so the quotient group $G/Z$ is a smaller p-group. By induction it has a central seriesLet $H_i$ be the inverse image of ${\overline{H}}_i$ under the quotient map, so $H_0=Z$ and $H_s=G$. Centrality in the quotient gives $[H_i,G]\le H_{i-1}$ for $i\ge 1$: the image of every such group commutator lies in ${\overline{H}}_{i-1}$. Also $[Z,G]=1$. Thereforeis a central series of $G$. This proves every finite p-group is nilpotent, including the case $Z=G$, when the quotient is trivial and the series is simply $1\le G$.

A central series is a finite chain of normal subgroupswith $G_i/G_{i-1}\le Z(G/G_{i-1})$ for every $i\ge 1$. Equivalently, writing $[x,y]=x^{-1}{y}^{-1}xy$ and $[A,B]$ for the subgroup generated by these group commutators, the condition is $[G_i,G]\le G_{i-1}$. Centrality is relative to the whole quotient group, not just to the next subgroup in the chain.

A group is nilpotent if it admits a finite central series. The trivial group admits a series of length zero. For a nontrivial nilpotent group, the minimum possible length is its nilpotency class.

Let $1=A_0\le \cdots \le A_r=A$ and $1=B_0\le \cdots \le B_s=B$ be central series. Pad the shorter chain by repeating its terminal group, so both have length $t=\max (r,s)$. In the direct product of groups, group commutators are coordinatewise, henceThe equality holds because commutators supported in either coordinate generate the corresponding factor independently. Thereforeis a central series, proving that a direct product of two nilpotent groups is nilpotent. More precisely, the coordinatewise recurrence for the lower central series gives ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$, so the nilpotency class of a direct product is the maximum of the two factor classes.

Define the lower central series by ${\Gamma }_1(G)=G$ and ${\Gamma }_{j+1}(G)=[{\Gamma }_j(G),G]$. Its terms are characteristic normal subgroups and form a descending chain. Define the upper central series by $Z_0(G)=1$ andEquivalently, $Z_{i+1}(G)=\{x\in G:[x,G]\subseteq Z_i(G)\}$. These are characteristic normal subgroups in an ascending chain, and successive factors are central in the appropriate quotient.

For any length-$r$ central series $1=G_0\le \cdots \le G_r=G$, prove the two comparisons independently. Starting at ${\Gamma }_1(G)=G_r$, if ${\Gamma }_{r-i+1}(G)\le G_i$, thenDescending induction gives the lower containment. Starting at $G_0=Z_0(G)=1$, if $G_{i-1}\le Z_{i-1}(G)$, then $[G_i,G]\le G_{i-1}\le Z_{i-1}(G)$, which by definition gives $G_i\le Z_i(G)$. Thus the central-series comparison theorem yieldsIn particular, a nilpotent group satisfies ${\Gamma }_{r+1}(G)=1$ and $Z_r(G)=G$.

Conversely, if ${\Gamma }_n(G)=1$, reverse the finite lower chain to obtain $1={\Gamma }_n\le {\Gamma }_{n-1}\le \cdots \le {\Gamma }_1=G$. The defining commutator recurrence makes it a central series. If $Z_n(G)=G$, the finite upper chain is already a central series. This provesFinally, for an integer $c>0$, termination at ${\Gamma }_{c+1}=1$ gives a central series of length $c$ and hence $Z_c=G$ by the comparison. Conversely, $Z_c=G$ gives a central series of length $c$ and hence ${\Gamma }_{c+1}=1$. ThereforeRepeated terms cause no problem; the least such $c$ for a nontrivial group is its nilpotency class.

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

The Kolchin theorem conjugates a faithful representation of a unipotent algebraic group $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication givesSince $U_n^{(n)}=1$, this filtration is a finite central series, so $U_n$ and every subgroup of it are nilpotent groups. Hence $U$ is nilpotent.

The converse fails: ${\mathbb{G}}_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.